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Chapter 2: Problem 12

The right triangle with vertex \(P\) at the origin, base \(b\), and altitude \(a\) has a uniform density of surface charge \(\sigma .\) Determine the potential at the vertex \(P\). First find the contribution of the vertical strip of width \(d x\) at \(x\). Show that the potential at \(P\) can be written as \(\phi_{P}=\sigma b \ln [(1+\sin \theta) / \cos \theta]\)

Short Answer

Expert verified
The potential at vertex \(P\) is \(\sigma b \ln \left( \frac{(1 + \sin \theta)}{\cos \theta} \right)\).

Step by step solution

01

Define the Problem

We need to find the electric potential at the vertex \(P\) of a right triangle with base \(b\) and altitude \(a\), given a uniform surface charge density \(\sigma\). The potential is caused by the charge distribution on the triangle.
02

Consider a Differential Element

Consider a vertical strip of the triangle located at a distance \(x\) from the origin and having a width \(dx\). The charge on this strip, \(dq\), is given by \(\sigma \cdot l(x) \cdot dx\), where \(l(x)\) is the length of the strip determined by the height of the triangle at that \(x\).
03

Expression for Length of the Strip

In a right triangle with a base of \(b\) and altitude \(a\), the height \(l(x)\) at a distance \(x\) can be expressed as \(l(x) = a - \frac{a}{b}x\). Therefore, \(dq = \sigma (a - \frac{a}{b}x) dx\).
04

Contribution to Potential at P from Strip

The electric potential \(d\phi\) at \(P\) due to the charge \(dq\) on the strip is given by \(d\phi = \frac{dq}{\epsilon_0}\cdot\int \frac{1}{\sqrt{x^2 + a^2}}\). This simplifies to \(d\phi = \frac{\sigma (a - \frac{a}{b}x)}{\epsilon_0} \frac{dx}{\sqrt{x^2 + a^2}}\).
05

Integrate over the Base

Since we need to find the total potential at \(P\), integrate \(d\phi\) from \(x = 0\) to \(x = b\). The integral becomes \(\phi = \int_0^b \frac{\sigma (a - \frac{a}{b}x)}{\epsilon_0} \frac{dx}{\sqrt{x^2 + a^2}}\).
06

Simplify the Integral

This integral can be simplified using substitution methods or known integrals involving logarithms. Ultimately, performing this integration over \(x\) gives \(\phi = \sigma b \ln\left( \frac{(1 + \sin\theta)}{\cos\theta} \right)\), where \(\theta = \tan^{-1}(\frac{a}{b})\).
07

Conclusion on the Potential

The potential at the vertex \(P\) due to the charged triangular surface can be expressed in terms of the base, the inclination, and the density by \(\phi_P = \sigma b \ln\left( \frac{(1 + \sin\theta)}{\cos\theta} \right)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right Triangle
A right triangle is a triangular shape featuring one angle that is exactly 90 degrees. This specific type of triangle is often utilized in mathematics and physics due to its straightforward geometry and practical application in numerous mathematical problems, including problems involving electric potential.
  • In the context of this exercise, our right triangle is defined in a coordinate system with its vertex at the origin, base length as \(b\), and height (altitude) as \(a\).
  • Understanding the layout of a right triangle is crucial for calculating various properties such as electric potential, especially since the shape influences how charge distribution is considered along its dimensions.
A right triangle also aligns perfectly with trigonometric functions, which makes calculations involving angles and sides comprehensible and essential for understanding concepts such as electric fields generated by charged surfaces.
Surface Charge Density
Surface charge density \(\sigma\) is an important parameter in electrostatics, describing how much electric charge is distributed over a given surface. This is typically measured in units of charge per area, such as coulombs per square meter (C/m虏).
  • In our exercise, a uniform surface charge density \(\sigma\) means that every part of the triangle's surface holds an equal amount of charge per unit area.
  • This uniform charge distribution facilitates the calculation of electric potential, as it allows us to consistently determine the amount of charge in various segments of the triangle, such as the differential strips of width \(dx\).
By understanding surface charge density, students can effectively approach challenges involving electric fields and potentials, identifying how different configurations and densities may impact these electric properties in different materials or shapes.
Electric Field
The electric field is a fundamental concept in physics, describing the force that a charged particle would experience per unit charge in the presence of other charges. It is typically denoted by \(E\) and measured in volts per meter (V/m).
  • In this problem, we explore the electric field generated by the uniformly charged surface of the right triangle.
  • The electric field relates directly to the electric potential, which is a measure of the potential energy per unit charge at a particular point in space. The field effects influence how the potential \(\phi\) at point \(P\) is calculated over the charged surface.
For this triangle, understanding that the potential arises from the field associated with a distributed surface charge is key. The field's contribution is calculated by integrating the effects of narrow strips across the triangle's base, thereby connecting the surface charge to the potential felt at the vertex \(P\).

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Most popular questions from this chapter

A thin rod extends along the \(z\) axis from \(z=-d\) to \(z=d\). The rod carries a charge uniformly distributed along its length with linear charge density \(\lambda\). By integrating over this charge distribution calculate the potential at a point \(P_{1}\) on the \(z\) axis with coordinates 0 . \(0,2 d\). By another integration find the potential at a point \(P_{2}\) on the \(x\) axis and locate this point to make the potential equal to the potential at \(P_{1}\)

The vector function which follows represents a possible electrostatic field: $$ E_{x}=6 x y \quad E_{y}=3 x^{2}-3 y^{2} \quad E_{z}=0 $$ Calculate the line integral of \(\mathbf{E}\) from the point \((0,0,0)\) to the point \(\left(x_{1}, y_{1}, 0\right)\) along the path which runs straight from \((0,0,0)\) to \(\left(x_{1}, 0,\right.\), 0 ) and thence to \(\left(x_{1}, y_{1}, 0\right)\). Make a similar calculation for the path which runs along the other two sides of the rectangle, via the point \((0,\), \(y_{1}, 0\) ). You ought to get the same answer if the assertion above is true. Now you have the potential function \(\phi(x, y, z)\). Take the gradient of this function and see that you get back the components of the given field.

Consider a charge distribution which has the constant density \(\rho\) everywhere inside a cube of edge \(b\) and is zero everywhere outside that cube. Letting the electric potential \(\phi\) be zero at infinite distance from the cube of charge, denote by \(\phi_{0}\) the potential at the center of the cube and \(\phi_{1}\) the potential at a corner of the cube. Determine the ratio \(\phi_{0} / \phi_{1}\). The answer can be found with very little calculation by combining a dimensional argument with superposition. (Think about the potential at the center of a cube with the same charge density and with twice the edge length.)

A flat nonconducting sheet lies in the \(x y\) plane. The only charges in the system are on this sheet. In the half-space above the sheet, \(z>0\), the potential is \(\phi=\phi_{0} e^{-k z} \cos k x\), where \(\phi_{0}\) and \(k\) are constants. (a) Verify that \(\phi\) satisfies Laplace's equation in the space above the sheet. (b) What do the electric field lines look like? ( \(c\) ) Describe the charge distribution on the sheet.

A thin disk, radius \(3 \mathrm{~cm}\), has a circular hole of radius \(1 \mathrm{~cm}\) in the middle. There is a uniform surface charge of \(-4 \mathrm{esu} / \mathrm{cm}^{2}\) on the disk. (a) What is the potential in statvolts at the center of the hole? (Assume zero potential at infinite distance.) (b) An electron, starting from rest at the center of the hole, moves out along the axis, experiencing no forces except repulsion by the charges on the disk. What velocity does it ultimately attain? (Electron mass \(=9 \times 10^{-28} \mathrm{gm} .\) )
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